Derivative of the square root of x?

Question

I know it's 1/(2squareroot(x)).
But how do I show that using the limit theorem?
Like with x^2: lim h->o ((x+h)^2 - x^2)/h
= (x^2 + h^2 + 2xh - x^2)/h
= (h^2 + 2xh)/h
=h + 2xh
=2x

I'm completely stumped by the square root aspect considering the power is 1/2, not 2, or 3.

Help!

Answer 1

lim h->0 (sqrt(x+h) - sqrt(x)) / h

** When you encounter equations like this with square roots, you should try this method below, applying the concept where :
a^2 - b^2 = (a+b)(a-b).

Think of your sqrt(x+h) as 'a' and sqrt(x) as 'b', so the equation you have now is (a-b). When you multiply it with (a+b), you'll get to square your 'a' and 'b'.. (sqrt(x))^2 is nothing but x.

So in this case, multiply the equation with

(sqrt(x+h) + sqrt(x))
--------------------------
(sqrt(x+h) + sqrt(x))

(which is essentially multiplying it by 1), you will get:

(sqrt(x+h))^2 - (sqrt(x))^2
---------------------------------
h (sqrt(x+h) + sqrt(x))

(x + h) - x
----------------------------
h (sqrt(x+h) + sqrt(x))

h
----------------------------
h (sqrt(x+h) + sqrt(x))

= 1 / (sqrt(x+h) + sqrt(x))

If h is approaching 0, (substitute h=0 into the equation) you get :

1 / (sqrt(x) + sqrt(x))

= 1 / 2*sqrt(x).

Answer 2

Multiply numerator and denominator by (√(x+h) + √x) and go from there. You'll end up with 1/(√(x+h) + √x) which becomes 1/(2√x) as h -> 0.

Doug

Answer 3

Write out the def:

lim (f(x+h)-f(x))/h as h->0

and use Taylor expansion for sqrt.

This might not be valid to use depending on the level of the class.

Answer 4

Man, your approach using limit theorem is not likely productive but let’s try!
dy=y(x+dx)-y(x)=sqrt(x+dx)-sqrt(x); now multiply and divide by sqrt(x+dx)+sqrt(x).
dy=[sqrt(x+dx)-sqrt(x)]*[sqrt(x+dx)+sqrt(x)]/[sqrt(x+dx)+sqrt(x)], or
dy=[x+dx-x]/ [sqrt(x+dx)+sqrt(x)]=dx/[sqrt(x+dx)+sqrt(x)], isn’t it?
Now dy/dx=1/[sqrt(x+dx)+sqrt(x)], let dx->0, then dy/dx=1/(2*sqrt(x)), that’s it!
It’ not always easy as that. So there is another theorem:
dy/dx=1/(dx/dy), i.e. u know, that y^2=x, then dx/dy=2y or dx/dy=2*sqrt(x)
thus dx/dy=1/(2*sqrt(x)).

Answer 5

[(x + h)^(0.5) - x^(0.5)] / h

Multiply thing by [(x + h)^(0.5) + x^(0.5)]/[(x + h)^(0.5) + x^(0.5)] = 1

=[(x + h) - x] / (h * [(x + h)^(0.5) + x^(0.5)])
=(h) / (h * [(x + h)^(0.5) + x^(0.5)])
=1 / [(x + h)^(0.5) + x^(0.5)]

Finally take the limit as 'h' goes to zero

=1 / (x^(0.5) + x^(0.5))
=1 / [2*(x^(0.5))]